Monopole-dipole expansion with an explicit remainder for compactly supported $L^1$ densities

Fix $y\in B_a$ and define $w(s):=x-sy$ for $s\in \mathbb{R}$, the open set $J_y:=\{s\in \mathbb{R}:w(s)\mathrm{\ne }0\}$, and $h:J_y\to \mathbb{R}$ by $h(s):=\parallel w(s){\parallel }_2^{-1}$. For $0\le s\le 1$ one has $\parallel w(s){\parallel }_2\ge r-s\parallel y{\parallel }_2\ge r-a\ge r\mathrm{/}2>0$ by [F1], so $[0,1]\subseteq J_y$ and $h$ is of class $C^2$ near $[0,1]$. To justify the root regularity, for $u>0$ rationalization gives $\mathrm{\mid }\sqrt{u+\eta }-\sqrt{u}\mathrm{\mid }\le \mathrm{\mid }\eta \mathrm{\mid }\mathrm{/}\sqrt{u}$ and $(\sqrt{u+\eta }-\sqrt{u})\mathrm{/}\eta =1\mathrm{/}(\sqrt{u+\eta }+\sqrt{u})$ for small nonzero $\eta$. Thus the root is continuous and has derivative $1\mathrm{/}(2\sqrt{u})$; the quotient rule makes this derivative continuously differentiable, so the reciprocal root is $C^2$. Composing with the positive polynomial $⟨w,w⟩$ and using [F2] now gives $h^{\mathrm{\prime }}(s)=⟨w(s),y⟩\parallel w(s){\parallel }_2^{-3}$ and $h^{\mathrm{\prime }\mathrm{\prime }}(s)=-\parallel y{\parallel }_2^2\parallel w(s){\parallel }_2^{-3}+3⟨w(s),y{⟩}^2\parallel w(s){\parallel }_2^{-5}$.

By Cauchy-Schwarz [F1], $\mid ⟨w(s),y⟩\mid \le \parallel w(s){\parallel }_2\parallel y{\parallel }_2$, so the terms of step 1.1 give $\mid h^{\mathrm{\prime }\mathrm{\prime }}(s)\mid \le \parallel y{\parallel }_2^2\parallel w(s){\parallel }_2^{-3}+3\parallel y{\parallel }_2^2\parallel w(s){\parallel }_2^{-3}=4\parallel y{\parallel }_2^2\parallel w(s){\parallel }_2^{-3}$. Since $\parallel w(s){\parallel }_2\ge r-a$ and $r-a\ge r\mathrm{/}2$ for $r\ge 2a$, this is at most $4\parallel y{\parallel }_2^2(r-a{)}^{-3}\le 32\parallel y{\parallel }_2^2{r}^{-3}$ for every $s\in [0,1]$.

Put $G(s):=(1-s)h^{\mathrm{\prime }}(s)+h(s)$ on $[0,1]$. $G$ is continuous on $[0,1]$ and differentiable on $(0,1)$ with $G^{\mathrm{\prime }}(s)=(1-s)h^{\mathrm{\prime }\mathrm{\prime }}(s)$, a continuous hence Riemann integrable function on $[0,1]$; the fundamental theorem [F3] applied to $G$ and $f:=(1-s)h^{\mathrm{\prime }\mathrm{\prime }}(s)$ therefore gives $R_y(x):={\int }_0^1(1-s)h^{\mathrm{\prime }\mathrm{\prime }}(s)ds=G(1)-G(0)=h(1)-h(0)-h^{\mathrm{\prime }}(0)$. Consequently $\parallel x-y{\parallel }_2^{-1}=h(1)=\parallel x{\parallel }_2^{-1}+⟨x,y⟩r^{-3}+R_y(x)$, with $h(0)=r^{-1}$ and $h^{\mathrm{\prime }}(0)=⟨x,y⟩r^{-3}$ by step 1.1; and monotonicity [F4] applied to $-\mathrm{\mid }{h}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }\le (1-s)h^{\mathrm{\prime }\mathrm{\prime }}\le \mathrm{\mid }{h}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid }$ together with step 2.1 gives $\mid R_y(x)\mid \le {\int }_0^1(1-s)\mid h^{\mathrm{\prime }\mathrm{\prime }}(s)\mid ds\le 32\parallel y{\parallel }_2^2{r}^{-3}{\int }_0^1(1-s)ds=16\parallel y{\parallel }_2^2{r}^{-3}\le 16a^2{r}^{-3}$.

The numbers $Q$, $p$, $\parallel \rho {\parallel }_1$ and the remainder are finite and well defined. Indeed, for each $k<3$ one has $\mid \rho y_k\mid \le a\mid \rho \mid$ on $B_a$ and $\rho y_k=0$ almost everywhere off $B_a$, so every scalar component $y_k\rho$ lies in $L^1({\lambda }_3)$ and $p\in {\mathbb{C}}^3$ is finite by [F5] and [F7]; for the remainder, the function obtained by extending $y\mapsto (\parallel x-y{\parallel }_2^{-1}-r^{-1}-⟨x,y⟩r^{-3})$ from $B_a$ by zero is Borel on ${\mathbb{R}}^3$, because the bracket is continuous on the open set $\{\parallel y{\parallel }_2<r\}\supseteq B_a$ by [F9] (the denominator $\parallel x-y{\parallel }_2\ge r-a>0$ is bounded away from zero on $B_a$) and ${\chi }_{B_a}$ is the indicator of a closed hence Borel set, and its product with $\rho$ is bounded in modulus by $\mid \rho \mid$ times the finite constant $C:=a^{-1}+r^{-1}+a{r}^{-2}$, since $\parallel x-y{\parallel }_2\ge r-a\ge a$ and $\mid ⟨x,y⟩\mid r^{-3}\le a{r}^{-2}$ on $B_a$; hence the product with $\rho$ lies in $L^1({\lambda }_3)$ by [F5] and [F7].

The stated expansion holds. Replacing $\rho$ by its representative ${\chi }_{B_a}\rho$ changes $Q$, $p$, $\parallel \rho {\parallel }_1$ and the left-hand integral by nothing, by [F8], so it suffices to prove the identity for ${\chi }_{B_a}\rho$, for which the pointwise identity of step 3.1 holds at every $y\in B_a$. For that representative, writing the left-hand integrand as $\rho (y){\chi }_{B_a}(y)K_x(y)$ and using ${\chi }_{B_a}={\chi }_{B_a}^2$, linearity [F5] of the Lebesgue integral on the space of integrable functions gives $\int \rho (y)\parallel x-y{\parallel }_2^{-1}d{\lambda }_3(y)=r^{-1}\int \rho d{\lambda }_3+r^{-3}{\sum }_{k<3}{x}_k\int y_k\rho (y)d{\lambda }_3+R(x)=Q\mathrm{/}r+⟨p,x⟩r^{-3}+R(x)$, with $R(x)$ as displayed in the Statement; the three summands are integrable by step 4.1 and the finite complex-linear combination is legitimate. The contraction is exactly ${\sum }_{k<3}{x}_k{p}_k$ in the Statement, so no conjugation or coordinate relabelling is used. For real $\rho$, each $p_k$ and $Q$ is real and the real kernel makes $R(x)$ real; the same identity and modulus estimate hold for complex $\rho$ by complex linearity.

The bound holds: by the triangle inequality [F6] and monotonicity [F7], $\mid R(x)\mid \le {\int }_{B_a}\mid \rho (y)\mid \mid \parallel x-y{\parallel }_2^{-1}-r^{-1}-⟨x,y⟩r^{-3}\mid d{\lambda }_3(y)\le 16a^2{r}^{-3}\int \mid \rho \mid d{\lambda }_3=16a^2\parallel \rho {\parallel }_1{r}^{-3}$, the middle inequality being the pointwise bound $16a^2{r}^{-3}$ for the bracket on $B_a$ from step 3.1 and the final equality the definition of $\parallel \rho {\parallel }_1$. Hence the expansion and the explicit remainder bound hold for every $r\ge 2a$.